Page # 1

WAVE OPTICS

Coherent and Incoherent SourcesWhy do we not commonly see interference effects with visible light? With light from a source such as theSun, an incandescent bulb, or a fluorescent bulb, we do not see regions of constructive and destructiveinterference; rather, the intensity at any point is the sum of the intensities due to the individual waves.Light from anyone of these sources is, at the atomic level, emitted by a vast number of independentsources. Waves from independent sources are incoherent; they do not maintain a fixed phase relation-ship with each other. We cannot accurately predict the phase (for instance, whether the wave is at amaximum or at a zero) at one point given the phase at another point. Incoherent waves have rapidlyfluctuating phase relationships. The result is an averaging out of interference effects, so that the totalintensity (or power per unit area) is just the sum of the intensities of the individual waves. Only thesuperposition of coherent waves produces interference. Coherent waves must be locked in with a fixedphase relationship. Coherent and incoherent waves are idealized extremes; all real waves fall some-where between the extremes. The light emitted by a laser can be highly coherent-two points in the beamcan be coherent even if sep arated by as much as several kilometers. Light from a distant point source(such as a star other than the Sun) has some degree of coherence.

Thomas Young (1773-1829) performed the first visible-light interference experiments using a clever technique toobtain two coherent light sources from a single source. When a single narrow slit is illuminated, the lightwave that passes through the slit diffracts or spreads out. The single slit acts as a single coherent sourceto illuminate two other slits. These two other slits then act as sources of coherent light for interference.

INTENSITY OF TWO SOURCE INTERFERENCEWe now obtain an expression for the distribution of intensity of two coherent sources that are in phase.The wave function in this case is the electric field. We assume that the slits are narrow enough fordiffraction to spread light from each slit uniformly over the screen. Thus, the amplitude of the fields anypoint on the screen will be equal. At a given point of the screen the fields due to S1 and S2 are

E1 = E0 win (t) ; E2 = E0 sin (t + )where the phase difference depends on the path difference = r2 r1. Since one wavelength corresponding to a phase change of 2, a distance corresponds to a phase change given by / 2 = / . If the screen is far from the slits, d sin (see the ref. figure for YDSE), therefore

= 2

=

sind2

The resultant field is found from the principle of superposition :E = E1 + E2 = E0 sin (t) + E0 sin (t + )

TEACHING NOTESTEACHING NOTES

Page # 2

By using the trigonometric identity sin A + sin B = 2 sin [(A + B)/2] cos [(A B)/2], we obtain

E = 2E0 cos

2 sin

2t

The amplitude of the resultant wave is 2E0 cos (/ 2). The intensity of a wave is proportional to thesquare of the amplitude, so from equation wave have

I = 4I0 cos2 cos2

2where I0 E0

2 is the intensity due to a single source. This function is plotted in fig. of pt (iv). The maximaoccur when = 0, 2, 4, ... = 2m. At these points I = 4I0 ; that is the intensity is four times that of asingle source. The minima (I = 0) occur when = , 3, 5, ....... = (2m + 1).

(i) If amplitudes of waves arriving at point P on the screen are different then resultant intensity is given by

I = I1 + I2 + 21II2 cos

Also, Imax = 221 II , when cos = 1Imin = 221 II , when cos = 1

(ii)min

max

II

=

2

21

21

IIII

=

2

21

21

AAAA

= 2

1r1r

where r = 2

1

2

1

II

AA

(iii) The phenomenon of interference is based on conservation of energy. There is no destruction of energy inthe interference phenomenon. The energy which apparently disappears at the minima, has actually beentransferred to the maxima where the intensity is greater than that produced by the two beams actingseparately.

Iav = 21

2

021212

0

2

0 IIcosII2II(21

d

Id

2

0

0dcos

as the average value of intensity is equal to the sum of individual intensities, therefore the energy is notdestroyed but merely redistributed in the interference pattern.

Page # 3

(iv) All maxima are equally spaced and equally bright. This is true for minima as well. Also interferencemaxima and minima are alternate. The intensity distribution in interference pattern is shown in figure.

O /2 In

tens

ity (I

) 221max III

21av III 2

21min III

Path difference ( x)

(iv) Path difference (x) and phase difference () are related as given below : path difference = 2 phase difference

or x = 2

YOUNGS DOUBLE SLIT EXPERIMENT (YDSE)The experiment set up for Youngs double slit experiment is shown in figure. Light after passing througha pin hole S is allowed to fall on thin slits S1 and S2 placed symmetrically w.r.t. S. A screen isplaced at a distance D from S1 and S2.

y

Q O

D

S1

S2

P

d

S1

S2x

MonochromaticSource Screen

Geometric construction for describing Youngs double-slit experiment

S

(D >> d)

Let P be the point, at which we want to investigate the intensity. Two rays S1P and S2P starting from S1and S2 reach P and interfere with each other.If x is the path difference between two rays,

x = S2P S1P d sin Ddy

DytansinanglesmallFor

This equation assumes that S1P and S2P are parallel, which is approximately true because D is muchgreater than d.

(a) Maxima : Point P will be a bright spot if the path difference x is integral multiple of .

yn = dnD

where, n = 0, 1, 2, 3, .......

Thus, bright spots are obtained at distances, 0, dD

, dD2

, dD3

....... from O.

(b) Minima : Point P will be a dark spot if the path difference x is an odd multiple of 2

.

Page # 4

i.e., if Dyd

= 2)1n2(

yn = d2D)1n2(

where, n = 0, 1, 2, 3, ........

Thus, dark spots are obtained at distances, d2D5,

d2D3,

d2D

....... from O.

(c) Fringe width ()It is the distance between two consecutive bright or dark fringes.

Let yn and yn 1 respectively, be the distances of nth and (n 1)th bright fringe from O,

= [yn yn1] = n dD

(n 1) dD

= dD

(n n + 1)

or = dD

Similarly, it can be proved that distance between two consecutive dark fringes, is given by

= dD

= = D / dHence, the bright and dark fringes are equally spaced.

[Home Work : HCV : 1 to 11Sheet Ex.1 : 1 Ex. 3 : 1, 6Q. Bank : 1, 3 AR : 2 ]

Remarks :(i) If whole apparatus is immersed in liquid of refractive index then,

= dD

i.e., fringes width decreases

(ii) Some times in numerical problems, angular fringe width () is given which is defined as angular separationbetween two consecutive maxima or minima

= dD

In medium, other than air or vacuum,

= d

Page # 5

(iii) x = Dyd

is valid when angular position of maxima or minima is less than 6

. However x = d sin is

valid for larger values of provided d

Page # 6

SHAPE OF INTERFERENCE FRINGES IN YDSEWe discuss the shape of fringes when two pinholes are used instead of the two slits in YDSE.Fringes are locus of points which move in such a way that its path difference from the two slits remainsconstant.

y

S1

S2

O x

Px

S2P S1P = D = constant ........(1)

If = 2

, the fringe represents 1st minima.

S2

S1

Y

X

= 3 = 2 =

= = =

If = 23

it represents 2nd minima

If = 0 it represents central maxima,If = , it represents 1st maxima etc.Equation (1) represents a hyperbola with its two foci at S1 and S2The interference pattern which we get on screen is the section of hyperboloid of revolution when werevolve the hyperbola about the axis S1S2.

(A) If the screen is to the X-axis, i.e. in the YZ plane, as is generally the case, fringes are hyperbolic witha straight central section.

(B) If the screen is in the XY plane, again fringes are hyperbolic.(C) If screen is to Y-axis (along S1S2) i.e. in the XZ plane, fringes are concentric circles with center on the

axis S1S2 ; the central fringe is bright if S1S2 = n and dark if S1S2 = (2n 1) 2

.

Z

A

Y

Y

B

X

X

Y

C[Home Work :HCV : 29 to 32Sheet Ex.1 : 2 to 6 Ex.2 : 1 Ex.3 : 7, 12, 13Q Bank MCQ : 6, 7, 8, 29 ]

{tell students to attempt from the book or can be given as HW}HCV-Ex. 33

Page # 7

Ex. Consider the situation shown in figure. The two slits S1 and S2 placed symmetrically around the centralline are illuminated by a monochromatic light of wavelength . The separation between the slits is d. Thelight transmitted by the slits falls on a screen 1 placed at a distance D from the slits. The slit S3 is at thecentral line and the slits S4 is at a distance z from S3. Another screen 2 is placed a further distance Daway from 1. Find the ratio of the maximum to minimum intensity observed on 2 if z is equal to

S1

S2

S4

S3

1

z

2D D

d

(a) d2D

(b) d4D

[Sol.(a) Let I is intensity due to slits S1 and S2 on screen S1. Further, intensity at any point on screen 1 is givenby

IP = 4I cos2

2At slit S3, = 0

3SI = 4I

At slit S4, x = Ddz

= 2

= 4SI = 0

Now on screen 2 Imax = 2SS 43 II = 4IImin = 2SS 43 II = 4I

min

max

II

= 1 Ans

(b) z = d4D

3SI = 4I

At slit S4, x = Ddz

= 4

= 2

4

= 2

4SI = 4I cos2

4 = 2I

Page # 8

Imax = 2SS 43 II = 2I2I4 = I (2 + 2 )2

Similarly, Imin = 2SS 43 II = 2I2I4 = I 222

min

max

II

= 2

2222

Ans.

HCV-Ex. 28Ex. Figure shows three equidistant slits being illuminated by a monochromatic parallel beam of light. Let

BP0 AP0 = /3 and D >> . (a) Show that in this case d = 3/D2 . (b) Show that the intensity at P0is three times the intensity due to any of the three slits individually.

d

d

DA

B

C

P0

Sol.

d

d

DA

B

C

P0

BP0 AP0 = 3

or d sin = 3

or d tan = 3

(For small angle tan sin )

d

D2/d

= 3

or d = 3F2

(b) xA/B = path difference between waves coming from A and B = 3

A/B = phase difference

= 2xA/B = 3

2

Similarly, xB/C = d sin

Page # 9

= d

D2/d3

= D2d3 2

= 1

B/C = 2Now, phase diagram of the waves arriving at P0 is as shown below :

AA

A

120

2A

A

120

Amplitude of resultant wave is given by

A = 120cos)A2)(A(2)A2(A 22 = 3 AAs intensity (I) A2 Intensity at P0 will be three times the intensity due to any of the three slits individually.

[Home Work : HCV : 20, 25, 26, 27, 28, 33, 34Sheet Ex-1 : 11, 12, 13 Ex.3 : 4, 14, 18Q. Bank : Single correct 4, 5, 9, 10, 11, 13, 14, 15, 16, 24

MCQ 2, 3, 4, 10, 11, 12, 13 AR 3 ]

GEOMETRICAL PATH AND OPTICAL PATHActual distance travelled by light in a medium is called geometrical path (x). Consider a light wavegiven by the equation.

E = E0 sin (t kx + )

If the light travels by x, its phase changes by kx = vx, where , the frequency of light does not

depend on the medium, but v, the speed of light depends on the medium as v = c

.

Consequently, change in phase,

= kx = C

(x)

It is clear that a wave travelling a distance x in a medium of refractive index suffers the same phasechange as when it travels a distance x in vacuum. i.e. a path length of x in medium of refractive index is equivalent to a path length of x in vacuum.

The quantity x is called the optical path length of light, xopt. And in terms of optical path length, phasedifference would be given by,

= Cxopt = 0

2

xopt

where 0 = wavelength of light in vacuum.However in terms of the geometrical path length x,

= C

(x) = 2x

where = wavelength of light in the medium

0 .

Page # 10

1. Displacement of fringe on introduction of a glass slab in the path of the light coming out of the slits.

D

S2

S1

P

O

O

d

t

On intrudction of the thin glass-slab of thickness t and refractive index , the optical path of the ray S1Pincreases by t( 1). Now the path difference between waves coming from S1 and S2 at any point P is

p = S2P (S1P + t( 1))= (S2P S1P) t( 1)

p = d sin t ( 1) if d

Page # 11

The phase change on reflectionA ray of light is incident on air-water interface ; let the amplitude reflection and transmission coefficientsbe r1 and t1 respectively. The amplitudes of reflected and transmitted waves are ar1 and at1 respectively.From the principle of reversibility of light, the system retraces its whole previous motion. The wave ofamplitude ar1 gives a reflected wave of amplitude ar1t1. The wave of amplitude at1

r ,t1 2

r ,t2 1

at t1 2

gives a reflected wave of amplitude ar2t1 and transmitted wave amplitude at1t2.So ar1

2 + at1t2 = at1t2 = 1 t1

2 ... (1)Further, the waves of amplitudes at1r2 and ar1t1 must cancel each other.

at1r2 + ar1t1 = 0r2 = r1 ... (2)

equation shows a difference of phase of between the two cases ; a reversal of sign means a displacementin the opposite sense. If there is no change of phase on reflection from above, there must be a phasechange of from below and vice-versa.When light gets reflected from a denser medium there is an abrupt phase change of ; no phase changeoccurs when reflection takes place from rarer medium.

Thin film interferenceWhen light passes the boundary between two transparent media, some light is reflected at the boundary.As shown in the figure some light is reflected from first interface and some from second interface. If weconsider a monochromatic incident light the two reflected waves are also monochromatic and coherentbecause they arise from the same monochromatic incident light wave via amplitude division. Thesewaves interfere, since they are superposed along the same normal line.The phase difference between two interfering waves is due to :(1) Optical path difference (due to distances travelled),(2) Reflection from a denser medium.

The second factor is irrelevant for reflection at rarer medium.Three situations may arise :(1) Neither wave experiences a phase change upon reflection.(2) Both the waves suffer a phase change upon reflection.In either of these two cases the phase change due to reflection is irrelevant ; no difference in phase resultsdue to reflection.In either of these cases phase change is determined solely from optical path difference.

Condition for constructive interference :

Page # 12

2t = mCondition for destructive interference :

2t =

21m

where m = 0, 1, 2, .......(3) One of the reflected waves experiences a phase change of radian upon reflection and the otherwaves does not.It is material which wave suffers a phase change ; the conclusions in the previous case are first reversed.

Condition for destructive interference :2t = m

Condition for constructive interference :

2t =

21m

where m = 0, 1, 2, .......

Ex. A light ray is incident normal to a thin layer of glass. Given the figure,what is the minimum thickness of the glass that gives the reflected lightan orangish color ( air = 600 nm)?(A) 50 nm (B*) 100 nm (C) 150 nm (D) 200 nm (E) 500 nm

[Sol. For reflected light to have orangish color, rays from A, C, E must be out of phase for l = 600 nmor = (2n + 1)

or 2gt (2n + 1) 2

i.e. t = g4

)1n2(

A C Et

DB

Transmitted light

(dense)(rare) water

air(rare)glass

Reflected light

or tmin = g4

= 100 nm ]

(OPTIONAL) INTERFERENCE DUE TO REFLECTED LIGHT :Consider a transparent film of thickness t and refractive index . A ray SA incident on the upper surfaceof the film is partly reflected along AR1 and partly refracted along AB. At B part of it is reflected alongBC and finally emerges out along CR2. The difference in path between the two rays. AR1 and CR2 iscalculated as given below :Let CN and BM be perpendicular to AR1 and AC. As the paths of the rays AR1 and CR2 beyond CNare equal. The path difference between them is

x = Path ABC in film Path AN in air= (AB + BC) AN = 2 AB AN

r

i C Air

R2R1N

i

S

A

B

rM

t

Air

T1

Page # 13

Now, AB = BC = BMAB

BM = BM sec r = t sec r

and AN = ACAN

AC = AC sin i = 2 AM sin i

= 2 BMAM

BM sin i = 2 (tan r) t

rsin

rsinisin

= 2t rcosrsin 2

= 2t sec r sin2 r

Then, = 2AB AN = 2t sec r 2 t sec r sin2 r= 2 t sec r (1 sin2 r) = 2 t cos r

The ray AR1 having suffered a reflection at the surface of denser medium undergoes a phase change or

path diff. of 2

.

At B the reflection takes place when the ray is going from a denser to rarer medium and there is no phasechange.Hence, the effective path difference between AR1 and CR2 is given by

Path Diff. (x) = 2 t cos r 2

(i) If the path difference x = n where n = 0, 1, 2, 3, 4 etc., constructive interference takes placeand the film appears bright.

2t cos r 2

= n

or 2t cos r = (2n + 1) 2

(ii) If the path difference x = (2n + 1) 2

where n = 0, 1, 2, ........... etc., destructive interference

takes place and the film appears dark.

2t cos r 2

= (2n + 1) 2

or 2t cos r = n

Remarks :(i) If the thickness of the film is very small as compared to the wavelength of light used, so that 2t cos r can

be neglected, then the total path difference between AR1 and CR2 will reduce to 2

. Thus two rays will

interfere destructively and darkness will result.

(ii) It should be remembered that the interference pattern will not be perfect because the intensities of the rayAR1 and CR2 will not be the same and their amplitudes are

Page # 14

Fringes of equal thicknessSoap bubbles and oil films on a road do not have uniform thickness of the film at any given pointdetermines whether the reflected light has a maximum or minimum intensity. When white light is used,each wavelength has its own fringe pattern. At a given point of the film, one wavelength may be enhancedand / or another wavelength suppressed. This is the source of the colors in soap bubbles an oil films onthe road.A wedge-shaped film of air may be produced by placing a sheet ofpaper or a hair between the ends of two glass plates, as in fig. With flatplates, one sees a series of bright and dark bands, each characteristic ofa particular thickness. If the plates are not flat, the fringes are not straight; each is locus of points with the same thickness. If one plate is knownto be flat, the fringes display the irregularities of the other, as shown infigure. The pattern shows where the plate needs to be polished for it tobe made optically flat.

Ex. A wedge-shaped film of air is produced by placing a fine wire of diameter D between the ends of two flatglass plates of length L = 20 cm, as in fig. When the air film is illuminated with light of wavelength = 550nm, there are 12 dark fringes per centimeter. Find D.

Sol. A indicated in fig. only one of the reflected ray suffers a phase inversion. At the thin end of the wedge,where the thickness is less than /4, the two rays interfere destructively. This region is dark in thereflected light. The condition for destructive interference in the reflected light is

2t = m m = 0, 1, 2, .....the change in thickness between adjacent dark fringes is t = / 2. The horizontal spacing betweenfringes d = 1/12 cm = 8.3 104 m. From figure we see that D / L = t / , so

= d2L

= m106.16)m2.0()m105.5(

4

7

Thus D = 6.6 105 m

Newtons RingsWhen a lens with a large radius of curvature is placed on a flat plate, as to fig. a thin film of air is formed.When the film is illuminated with monochromatic light, circular fringes, called newtons rings, can be withthe unaided eye or with a low power microscope (figure). An important feature of Newtons rings is thedark central spot. Newton tried polishing the surfaces to get rid of it, The dark spot was also initiallypuzzling to Young. It implied that the light wave suffers a phase inversion on reflection at a medium witha higher refractive index. Young tested this idea by placing oil of sassafras between a lens of crown glassand a plate of flint glass. The refractive index of the oil is between the values for these two glasses. Sinceboth reflections occur at a medium with a higher refractive index, they should both suffer a phase inversionand therefore be in phase be in phase. This is precisely what happened : The central spot became brightand undoubtedly gave Young much satisfaction.

Page # 15

Ex. In an experiment on Newtons rings the light has a wavelength of 600 nm. The lens has a refractive indexof 1.5 and a radius of curvature of 2.5 m. Find the radius of the 5th bright fringe.

Sol. If R is the radius of curvature of the lens, then from fig. we see that r2 = R2 (R t)2, where r is the radiusof a fringe and t is the thickness of the film. Since t is very small, we may drop terms in t2 to obtain.

r2 2Rt ... (i) In order to find r, we must first find t. The condition for a bright fringe is

2t = F21m

... (ii)

We note that n = 1 for the air film (the index for the glass is irrelevant) and that m = 4 for the fifth brightfringe. Thus, from (ii)

t = 2

)106()5.4( 7 = 1.35 106 m

Substituting this into (i), we find r = Rt2 = 2.6 103 m.

[ Home Work : HCV : 35, 36, 37, 38Sheet Ex.1 : 14 to 17Ex.2 : 8, 9Q.Bank : Single correct - 25, 26, 27

MCQ - 7 ]

HUYGENS CONSTRUCTION(This matter is from NCERT)Huygens, the Dutch physicist and astronomer of the seventeenth century, gave a beautiful geometricaldescription of wave propagation. We can guess that he must have seen water waves many times in thecanals of his native place Holland. A stick placed in water and oscillated up and down becomes a sourceof waves. Since the surface of water is two dimensional, the resulting wavefronts would be circlesinstead of spheres. At each point on such a circle, the water level moves up and down. Huygens idea isthat we can think of every such oscillating point on a wavefront as a new source of waves. According toHuygens principle, what we observe is the result of adding up the waves from all these different sources.These are called secondary wave or wavelets. Huygens principle is illustrated in (figure 10.1) in thesimple case of a plane wave.

(i) Consider a plane wave passing through a thin prism. Clearly, the portion of the incoming wavefrontwhich travels through the greatest thickness of glass has been delayed the most. Since light travels moreslowly in glass. This explains the tilt in the emerging wavefront.

(ii) Similarly, the central part of an incident plane wave traverses the thickest portion of a convex lens and isdelayed the most. The emerging wavefront has a depression at the centre. It is spherical and convergesto a focus.

Page # 16

(iii) A concave mirror produces a similar effect. The centre of the wavefront has to travel a greater distancebefore and after getting reflected, when compared to the edge. This again produces a converging sphericalwavefront.

(iv) Cocave lenses and convex mirror can be understood from time delay arguments in a similar manner. Oneinteresting property which is obvious from the pictures of wavefronts is that the total time taken from apoint on the object to the corresponding point on the image is the same measured along any rays goingthrough the centre are shorter. But because of the slower speed in glass, the time taken is the same as forrays travelling near the edge of the lens.

[Home Work :Sheet : Ex.3 : 3, 15, 16, 17Q. bank : single correct - 2, 12, 30 ]